/*
 * @Author: liusheng
 * @Date: 2022-07-23 18:04:02
 * @LastEditors: liusheng
 * @LastEditTime: 2022-07-23 18:34:12
 * @Description: 剑指 Offer II 099. 最小路径之和
 * email:liusheng613@126.com
 * Copyright (c) 2022 by liusheng/liusheng, All Rights Reserved. 
 * 
 剑指 Offer II 099. 最小路径之和
给定一个包含非负整数的 m x n 网格 grid ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。

说明：一个机器人每次只能向下或者向右移动一步。

 

示例 1：



输入：grid = [[1,3,1],[1,5,1],[4,2,1]]
输出：7
解释：因为路径 1→3→1→1→1 的总和最小。
示例 2：

输入：grid = [[1,2,3],[4,5,6]]
输出：12
 

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 200
0 <= grid[i][j] <= 100
 

注意：本题与主站 64 题相同： https://leetcode-cn.com/problems/minimum-path-sum/

通过次数13,999 提交次数19,030
 */

#include "header.h"

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int>> dp(m,vector<int>(n,0));

        dp[0][0] = grid[0][0];
        for (int i = 1; i < m; ++i)
        {
            dp[i][0] = dp[i-1][0] + grid[i][0];
        }

        for (int j = 1; j < n; ++j)
        {
            dp[0][j] = dp[0][j-1] + grid[0][j];
        }

        for (int i = 1; i < m; ++i)
        {
            for (int j = 1; j < n; ++j)
            {
                dp[i][j] = grid[i][j];
                dp[i][j] += dp[i][j-1] < dp[i-1][j] ? dp[i][j-1] : dp[i-1][j];
            }
        }

        return dp[m-1][n-1];
    }
};

/*
scroll array optmize
*/
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<int> dp(n,0);
        vector<int> predp = dp;

        dp[0] = grid[0][0];
        for (int i = 1; i < n; ++i)
        {
            dp[i] = dp[i-1] + grid[0][i];
        }

        for (int i = 1; i < m; ++i)
        {
            predp = dp;
            for (int j = 0; j < n; ++j)
            {
                dp[j] = grid[i][j];
                if (j > 0)
                {
                    dp[j] += dp[j-1] < predp[j] ? dp[j-1] : predp[j];
                }
                else
                {
                    dp[j] += predp[j];
                }
            }
        }

        return dp[n-1];
    }
};
